6. Chemical Equilibrium (15 marks total) Consider the equilibrium at 25°C: PCI(g

Question

6. Chemical Equilibrium (15 marks total) Consider the equilibrium at 25°C: PCI(g) PC,() + Ch(g) and the information: Substance PCls(g) -324.6 -286.3 0 398.9 -306.4 0 Ch(g) Compute -,Go (2 marks),A,onHo (2 marks) and equilibr marks) at 25°C. a) ium constant K (2 b) Calculate the equilibrium constant K (2 marks) and Apn G (2 marks) at 600K. c) Fill in the following table to calculate the degree of dissociation, a, of PCls(g) at 600K and a total pressure of 5.066 bar (5 marks). You may assume an activity coefficient of 1. PCIslg) PCI3(8) Cl2(g)- Total moles Initial Equilibrium

Explanation / Answer

Calculate enthalpy at 25C

Enthalpyor RXn = Enthalpy products - Enthalpy of reactants

Enthalpy of RXN = Enthalpy (PCl3 + Cl2) - Enthalpy (PCl5)

Hrxn = -306.4 - (-326.4) = 20 KJ

The same goes for Gibbs ( products - reactants )

Gibbs = -286.3 - (-324.6) = 38.3 KJ

for the equilibrium constant apply the next equation

G = - RT Ln K, G is gibbs, R is gas constant , T is temperature, R must be 8.314 J / mol K, T is 25 C or 298 K , Gibbs must be joules instead of Kj

38 300 = - 8.314 * 298 * Ln K

38300 = -2477.57 * Ln K

Ln k = 38300 / -2477.57 = - 15.45, K = exp (-15.45)

K = 1.95 x 10-7

B) we can apply vant hoff equation that matches K equilibrium with temperature

ln ( Kf / Ki) = H / R * (1 / 298 - 1 / Tf)

Kf is final equilibrium constant or the constant we want to know

Ki is the equilibrium constant we know

Tf is temperature at which the equilibrium constant we want to know exists

Ln Kf / Ki = 20000 / 8.314 * ( 1 / 298 - 1 / 600 )

Ln (Kf/ki) = 4.06

Kf/Ki = 58.155

Kf = 58.155 * 1.95x10-7 = 1.134 x 10-5

C) Now calculate the total number of moles for the equilibrium , use the ideal gas equation for this

PV = n R T, p is pressure, V is volume, n is moles, R is gas constant and T is temperature

5.066 bar must be changed to atm which are approximately 5 atm , if we assume a volume of 1 Liter then

n = PV / RT = 5 * 1 / (0.082 * 600) = 0.101 moles

If we make the ICE chart then

PCl5 ============== PCl3 + Cl2

I 0.101...............................0...............0

C -x...................................x.............x

E 0.101 -x ..........................x.............x

Kequilibrium is the relationship between products and reactants

K = [PCl3][Cl2] / [PCl5] = X2 / (0.101 - X) = 1.134 x 10-5

We can see that the equilibrium benefits the reactant since the K is very small so if we solve this equation we get a value of x of 0.001065 moles at equilibrium so

PCl5 ============== PCl3 + Cl2

I 0.101...............................0...............0

C -0.00106.......................0.00106........0.00106

E 0.101 -0.00106.........0.00106.........0.00106

E = 0.0999...............................0.00106.......0.00106 Total moles at equilibrium = 0.102 moles, as you can wee 1 mole produces 2 moles thats why we have "extra moles" on the products side

Xi 0.0999/0.102 = 0.979.......0.0104........0.0104

Pi .979*5= 4.89 atm................0.0521.........0.0521, Pi = Xi * Pt, just multiply the last row by 5 which is the total pressure in atm

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